The one-to-one functions $g$ and $h$ are defined as follows.
\[
\begin{array}{l}
g(x)=\frac{x+13}{11} \\
h=\{(-4,2),(3,-6),(7,8),(8,0)\}
\end{array}
\]
Find the following.
\[
\begin{array}{r}
g^{-1}(x)=\frac{x+13}{11} \\
\left(g^{-1} \circ g\right)(-2)=41 \\
h^{-1}(8)=11 \\
\hline
\end{array}
\]
Answer
\(h^{-1}(8)\) is the x-value such that \((x,8)\) is in the function \(h\). Looking at the definition of \(h\), we see that \(h^{-1}(8)=\boxed{7}\).
Step 1 ：Since \(g\) is the function that adds 13 and then divides by 11, \(g^{-1}\) is the function that multiplies by 11 and then subtracts 13. This lets us compute from inside out:
Step 2 ：\(g^{-1}(g(-2))\) becomes \(g^{-1}(1)\) after substituting \(-2\) into \(g(x)\) and simplifying.
Step 3 ：Then, \(g^{-1}(1)\) becomes \(11*1-13=\boxed{-2}\) after substituting \(1\) into \(g^{-1}(x)\) and simplifying.
Step 4 ：\(h^{-1}(8)\) is the x-value such that \((x,8)\) is in the function \(h\). Looking at the definition of \(h\), we see that \(h^{-1}(8)=\boxed{7}\).
